Question

Which of the following species is diamagnetic?

• A. $O_2$
• B. $O_2^{2-}$
• C. $O_2^{-}$
• D. $O_2^{+}$

Answer 1

Answer: (B)

If a species does not have any unpaired electron, then it is called diamagnetic and if a species has unpaired electron, then it is called paramagnetic.
(a) Oxygen molecule $\left(O_2\right)$ has following electronic configuration.
$O_{2}KK$
$(\sigma 2s{)}^2(\stackrel{\ast }{\sigma }2s{)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2$
${\left(\stackrel{\ast }{\pi }2p_x\right)}^1{\left(\stackrel{\ast }{\pi }2p_y\right)}^1$
This is paramagnetic due to presence of two unpaired electrons.
(b) Peroxide ion $\left(O_2^{2-}\right)$ has following electronic configuration.
$O_2^{2-}KK$
$(\sigma 2s{)}^2{\left(\stackrel{\ast }{\sigma }2s\right)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^2){\left(\pi 2p_y\right)}^2$
${\left(\stackrel{\ast }{\pi }2p_x\right)}^2{\left(\stackrel{\ast }{\pi }2p_y\right)}^2$
There is no unpaired electron in $O_2^{2-}$ ion.
So, it is diamagnetic.
(c) Superoxide ion $\left(O_2^{-}\right)$has following electronic configuration.
$O_2^{-}KK$
$(\sigma 2s{)}^2{\left(\stackrel{\ast }{\sigma }2s\right)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2$
${\left(\stackrel{\ast }{\pi }2p_x\right)}^2{\left(\stackrel{\ast }{\pi }2p_y\right)}^1$
There is one unpaired electron in $O_2^{-}.$So, it is paramagnetic.
(d) Oxygen molecule ion $\left(O_2^{+}\right)$has following electronic configuration.
$O_2^{+}KK$
$(\sigma 2s{)}^2(\stackrel{\ast }{\sigma }2s{)}^2{\left(\sigma 2p_z\right)}^2{\left(\pi 2p_x\right)}^2{\left(\pi 2p_y\right)}^2{\left(\stackrel{\ast }{\pi }2p_x\right)}^1$ There is one unpaired electron in $O_2^{+}$. So, it is paramagnetic.